Chapter 1.6, Problem 21E

To determine

To find: the formula for the inverse of the function $f\left(x\right)=1+\sqrt{2+3x}$.

Answer to Problem 21E

The formula for the inverse of the function $f\left(x\right)=1+\sqrt{2+3x}$ is $f^{-1}\left(x\right)=\frac{{\left(x-1\right)}^2-2}{3}$.

Explanation of Solution

Let the given function as $y=1+\sqrt{2+3x}$.

Solve this equation for x as follows.

$\begin{array}{l}y-1=\sqrt{2+3x}\\ {\left(y-1\right)}^2=2+3x\\ y^2-2y+1=2+3x\\ x=\frac{y^2-2y-1}{3}\end{array}$

Then, interchange x and y and obtain the inverse function, $y=\frac{x^2-2x-1}{3}$.

It can be expressed as $y=\frac{{\left(x-1\right)}^2-2}{3}$.

Thus, the required inverse function is $f^{-1}\left(x\right)=\frac{{\left(x-1\right)}^2-2}{3}$.